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The James Webb Space Telescope (JWST) is in a halo orbit around L2, at a sufficient radius around the Lagrange point that it is in perpetual sunlight. That allows it to have predictable solar power, but requires the very large and fragile five-layered heat shield.

What would the the design trade-offs have been for choosing a low-radius, perpetually shadowed L2 halo orbit and using a nuclear battery (such as a Pu-238 RTG, which is relatively easy to shield) for power?

Machavity
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pde
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    One thing I've gathered from a bit more research is that L2 itself is not in the Earth's full shadow (umbra); rather it's in the antumbra with only 93%^2 = 86% of sunlight shaded. What the most-shaded halo orbit is is a different question, but it's possible all of them have much more than 86% sunlight. – pde Dec 27 '21 at 20:52
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    Space 101: You never go nuclear when solar will do the job. Nuclear is more expensive and it is heavier to launch (which is also more expensive.) You go nuclear when you're going too far out for solar, or when you're putting a craft into a situation where it's going to be shadowed too much. – Loren Pechtel Dec 29 '21 at 00:57
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    @LorenPechtel yes good point, and uhoh also discussed it in the accepted answer, but in this instance the comparison is not solar vs nuclear, it's solar+five huge fragile umbrellas vs nuclear. – pde Dec 29 '21 at 02:21
  • He was talking about nukes in this case, I'm saying it's a general issue, not specific to the JWST. – Loren Pechtel Dec 29 '21 at 02:41
  • FWIW, L2 is almost in Earth's umbra so the sun may be partially blocked. Related: https://space.stackexchange.com/q/10355/21336 – Engineer Toast Dec 30 '21 at 17:06

4 Answers4

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There's no place to hide!

That would be great, but the problem is that there aren't any orbits like that.

The only way to keep the temperature of the telescope rock-solid steady is to keep it in constant sunlight and insulate the heck out of it.

Going in and out of eclipse would cause all kinds of thermal perturbations, and the L2 Lagrange point itself is just a little bit too far away to be in Earth's umbra, and probably (though I don't know for sure) station-keeping so close to L2 would be a lot harder than doing it in JWST's big halo orbit around L2.

In no particular order:

Either 20 nukes or no nukes!

What would the the design trade-offs have been for choosing a low-radius, perpetually shadowed L2 halo orbit and using a nuclear battery (such as a Pu-238 RTG, which is relatively easy to shield) for power?

Wikipedia says that JWST has about 2,000 watts of solar power.

Wikipedia says that NASA's flagship RTG, the Multi-mission radioisotope thermoelectric generator has an output in the beginning of about 2,000 watts of thermal power and only 125 watts electrical, after say 10 years that might be only 105 watts.

So you would need about twenty 45 kg RTGs, or another 900 kilograms to equal the power output of the much, much lighter solar panel, a far, far simpler technology you can almost order out of a (very fancy space) catalog these days.

uhoh
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    20 x 2000W heat would be a great deal of problem of its own, especially in a spacecraft that has to be kept dead cold in order to do its job. On the other hand, a 2000W solar panel is itself a heat shield. – fraxinus Dec 28 '21 at 13:10
  • @fraxinus I understood that there is a solar panel, which unfolds minutes after launch, and a separate heat shield that unfolds several days later. A regular solar panel might get pretty warm as it absorbs a lot of sunlight, it is black! So I guess that the solar panel does NOT function as a heat shield here – Roland Dec 28 '21 at 13:32
  • the difference between a solar panel and a radiator is what you point it at. – John Dvorak Dec 28 '21 at 13:33
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    A solar panel in its usual orientation casts a favorable shadow. – fraxinus Dec 28 '21 at 14:51
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    In addition on the last point, plutonium RTGs are not really a practically renewable resource right now. Most of the Pu-238 in the world was a byproduct of the production of weapons-grade plutonium (which isn’t really being produced much anymore). Getting any significant amount today would require isotope separation of plutonium that would be used as reactor fuel, which is not needed for it to be usable as reactor fuel and is not exactly easy to do anyway (and thus is not something anybody is really doing much of). – Austin Hemmelgarn Dec 28 '21 at 18:02
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    @Roland If you insulate the back of a solar panel, it functions as a heat shield. That makes it hotter and less efficient than radiating from the back, but in the JWST geometry, it can't radiate in that direction anyway. – John Doty Dec 28 '21 at 18:02
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    @JohnDoty even if you don't insulate the solar panel, it still radiates less than half of the initial amount of energy in the same general direction (and the thermal radiation is dispersed so at least some of it doesn't reach the target). This is no diffrent than the main 5-layer heat shield. The solar panel is a medicore and partial 6th layer. – fraxinus Dec 29 '21 at 11:53
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    @AustinHemmelgarn the usual route to PU238 is via neutron bombardment of neptunium 237 (made by neutron bombardment of U238) to get neptunium 238 then beta decay to get Pu238. This has the virtue that since the Np can be chemically separated from the Pu in the spent fuel from a uranium burning reactor, there is no need to attempt isotopic separation of Pu, you separate out the Np, load it into cartridges and blast those with neutrons in a reactor, the result will be pure Pu 238. You could (given suitable paperwork!) do this in almost any isotope production facility. – Dan Mills Dec 29 '21 at 17:40
  • @JohnDoty Never mind your comment is upvoted, but you are simplifying. Just read on the nasa site. The cooling shield is incredibly complex and takes 5 days to unfold and is much larger than the solar panel. How would you "just add" this to the back of the solar panel? I even guess that the solar panel must be unfolded first, to power the motors that unfold the cooling shield – Roland Dec 29 '21 at 20:51
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    @Roland The fact remains that a solar panel intercepts solar radiation, converts some of it to electricity, and radiates some or perhaps most of the remainder away from the spacecraft. So, the radiation field behind the panel has lower energy density than the field in front. It's not nearly enough of a shield for JWST, but it's a start. – John Doty Dec 29 '21 at 21:14
  • @JohnDoty Looks like you have never seen a picture of the JWST heat shield. We are discussing JWST here, not philosophical heat shields. The heat shield is 22 x 10 m = 220 m2. The solar panel consists of 5 panels of about 1.2 x 2 m =12 m2. And you never told us how you would "insulate the back of a solar panel". You make a start with "not nearly enough", I would say "not" – Roland Dec 30 '21 at 09:40
  • @fraxinus "shadow" . . . sure, a solar panel is an excellent light shield. Light. Visible light. Not IR. – Roland Dec 30 '21 at 09:47
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    @Roland The traditional Si solar panels get some of their electricity from IR. But it is not really important if it is IR or visible. Both IR and visible heat up whatever they touch. And, behind the solar panel (or anything without its own energy source for that matter), in a thermal equilibrium, the radiation is less intense over the whole spectrum. The 5-layer heat shield itself works pretty much the same. On Earth, one may get more IR in the shadow solely because our atmosphere filters a great deal of the solar IR in the first place. And it is still colder in the shadow. – fraxinus Dec 30 '21 at 10:11
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    @Roland Pictures? I've walked around a full-scale mockup. Yes, quantitatively the solar panel isn't much of a heat shield compared to the rest. It's still a heat shield. As for how you insulate the back of a solar panel, you do it with the usual methods. Plate it with gold and/or cover it with MLI. – John Doty Dec 30 '21 at 12:01
  • @JohnDoty Ahhhh, quantitatively no heat shield. It does not shield much heat, but it still IS a heat shield because you say so. Ok. Fine. Ceci n'est pas une pipe. – Roland Dec 30 '21 at 14:52
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    @Roland 5% isn't zero. A good thermal engineer would include the effect. In a space mission you save watts a milliwatt at a time in the design. Every tiny bit helps. – John Doty Dec 30 '21 at 14:57
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    I'm wondering how much heat load the gamma radiation from the massive plutonium reactor would cause in the telescope. I also wonder whether one would obtain permission to launch such a large amount of Plutonium at all, given the fallout of a launch failure. – Peter - Reinstate Monica Dec 31 '21 at 00:00
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    @Peter-ReinstateMonica Good thinking! The isotopes selected for RTGs don't emit gamma rays. They're alpha emitters, and those have ranges measured in microns inside. There is a tiny flux of neutrons though, and that certainly could still be a problem! Does curiosity's RTG generate neutrons as this NASA CheMin X-ray detection system webpage suggests? If so, how? However an actual reactor would be a nightmare! – uhoh Dec 31 '21 at 00:02
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    @uhoh I simply assumed there would be gamma and was too lazy to look it up. I'm being lazy again simply assuming that neutrons don't interact enough to constitute a heat problem. – Peter - Reinstate Monica Dec 31 '21 at 06:27
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JWST is receiving sunlight because it is relatively near to the sun. At Neptune, the amount of sunlight is much less.

As I understood, the heat shield is to shield the telescope from the heat of the sun. The telescope should cool down to just a few dozen kelvin in order to have low noise in the detector.

The solar panels are different from the heat shield. Solar panels are probably a much simpler source of energy than a nuclear option, and less dangerous in case the telescope would crash down to the earth.

Also, a nuclear power source would produce heat that needs to be shielded from the telescope that needs to be very cold for low noise observations.

Solar panels are black to absorb sunlight while a heat shield is reflective to absorb as little radiation as possible.

Solar panels unfolded 30 minutes after launch in order to communicate with ground: NASA site

The sunshield deployment starts at three days after launch, and ends at eight days after launch, in several steps, see: NASA deployment information

Please refer to nasa.gov for more details on James Webb.

Fred
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Roland
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    Re, "panels are black to absorb sunlight..." The best commercially available PV modules are around 30% efficient. They get hot when they absorb sunlight. Put your hand on a solar panel that's been soaking up full sun for half a day, and you'll see. And, that's on Earth where air carries a lot of heat away. In vacuum, the panels will only be cooled by radiation, and some of that will be directed toward the rest of the spacecraft. So sure, PV panels provide significant shade, but it's not cold-darkness-of-space shade. – Solomon Slow Dec 28 '21 at 15:32
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    @SolomonSlow Right. When you think about it, a solar panel may just pass half of the 70 % of not-absorbed radiation, especially the infrared part, to the spacecraft sitting in its "shade". The other half radiates back to the sun. – Roland Dec 28 '21 at 16:06
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    It cannot "pass" more than 50%, except by being transparent. And thansparent the solar panel is not (except maybe partially in x-ray and partially in longer than cm-scale radio waves) – fraxinus Dec 30 '21 at 10:22
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As others pointed out, the L2 point is not in full shadow. Even if it were, the telescope would still need the heat shield to protect it from infrared radiation coming from the Earth and Moon.

Aside from the weight problem (a nuclear power source is a lot heaver than an equivalent solar panel in this part of the solar system), parking right at the L2 point would make it harder for the telescope to "hear" signals from Earth. The Sun produces a lot of radio noise and if it is right behind the Earth, then picking out radio signals from the Earth will be a lot harder. (A lot like trying to see a traffic light when the Sun is right behind that light.) By using a halo orbit, the satellite can point the high gain antenna at the Earth and the Sun's noise will be off to the side. (Similar to blocking the sun with your hand to better see the traffic light.)

This answer and the other answers to that question should help with understanding some of the drawbacks to parking right at the L2 point.

Kyle A
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Your question suggests you haven't understood the importance of the nature of the L2 point.

Useful links to read more, if the brief description below isn't enough:

The L2 Lagrange point isn't an object. It’s a point in space where forces balance out to create a point that James Webb Space Telescope (JWST) will easily stay near. The "halo orbit" means it won't orbit exactly in an ellipse as usual, but in a slightly different varying curve, that needs correction now and then.

And that's the answer to your question.

.....at a sufficient radius around the Lagrange point that it is in perpetual sunlight. That allows it to have predictable solar power....

There isn't such a place as you visualise, there. L2 isn’t an object. Nowhere is "in the shadow", and you can't get into shadow by hugging it closer, because there's no solid object actually at L2, for JWST to orbit, like it would a planet or moon, and be in the shade of.

(Technically, L2 is one of five places where an object will broadly stay in the same position relative to Earth, with very minimal course correction being needed. Some matter may gather at or near L2, but no solid object in the sense you're thinking, able to provide shade and attract an orbiting satellite)

100 metres from L2 or half a million km from L2, it’s all the same thing - perpetual sunlight exposure (or more exactly it’s sliiiightly in Earth's shadow, so somewhat reduced sunlight) - but nothing actually there for it to hide behind.

Peter Mortensen
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Stilez
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    L2 is an unstable Lagrange point, so matter won't gather there. – TonyK Dec 28 '21 at 15:56
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    I think the OP was assuming that a "low-radius" orbit at the L2 would be able to stay perpetually within the Earth's shadow. – Michael Richardson Dec 28 '21 at 16:31
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    @Stilez: I'm sure the question was not about hiding behind an object at L2 but rather it was assumed L2 was in the umbra of Earth (which it almost is). – Felix Tritschler Dec 28 '21 at 16:38
  • @FelixTritschler - maybe but seems doubtful (perhaps). The question asks about a "low radius perpetually shadowed" halo orbit, which would mean total shadow if anything, not penumbral shadow. You could be right, but I think an OP with the background to consider shadow vs penumbra, and who knew more about L2, would have learned its far enough from earth to make total eclipse very unlikely, and/or intuit that even slight sunlight would be a problem at JWST temperatures. Hence my suspicion they see L2 as a place that can be hidden behind for shade, and my approach. Could be wrong though.... – Stilez Dec 28 '21 at 17:43
  • @TonyK - yes indeed, but there's absolutely no guarantee that nothing has recently ended up there, or in a chance short-lived halo orbit. Some dust or whatever. It may be unstable, but its still worth a mention IMO for accuracy, otherwise I'm implying no matter might gather at/near which would be wrong too. – Stilez Dec 28 '21 at 17:47
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    @FelixTritschler yes you gathered my assumptions correctly. But I realised a little after posting the question that L2 is outside the umbra. Also it seems like there might not be any truly "low-radius halo orbits", but three body dynamics are complicated, especially if you have a small amount of rocket fuel, and I haven't spent enough time to be sure of that... – pde Dec 29 '21 at 02:08
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    @Stilez: "umbra", not "penumbra" - the umbra is the fully shadowed region, i.e. total solar eclipse at these locations. And the "tip" of Earth's umbra is not that far away from L2: ~ 92% of the distance (to L2). – Felix Tritschler Dec 29 '21 at 11:48